Heat engineering calculation of floors located on the ground. Calculation of the heat loss of the floor over the ground in the coalbed Remarks and conclusions

According to SNiP 41-01-2003, the floors of the building floor, located on the ground and logs, are delimited into four zones-strips 2 m wide parallel to the outer walls (Fig. 2.1). When calculating heat loss through floors located on the ground or logs, the surface of the floor sections near the corner of the outer walls ( in I zone-lane ) is entered into the calculation twice (square 2x2 m).

Heat transfer resistance should be determined:

a) for non-insulated floors on the ground and walls located below ground level, with a thermal conductivity of l ³ 1.2 W / (m × ° C) in zones 2 m wide, parallel to the outer walls, taking R n.p . , (m 2 × ° С) / W, equal to:

2.1 - for zone I;

4.3 - for zone II;

8.6 - for zone III;

14.2 - for zone IV (for the remaining floor area);

b) for insulated floors on the ground and walls located below ground level, with thermal conductivity l c.s.< 1,2 Вт/(м×°С) утепляющего слоя толщиной d у.с. , м, принимая R u.p. , (m 2 × ° С) / W, according to the formula

c) thermal resistance to heat transfer of individual zones of floors on logs R l, (m 2 × ° C) / W, is determined by the formulas:

Zone I - ;

II zone - ;

III zone - ;

IV zone - ,

where,,, are the values ​​of thermal resistance to heat transfer of individual zones of non-insulated floors, (m 2 × ° С) / W, respectively, numerically equal to 2.1; 4.3; 8.6; 14.2; - the sum of the values ​​of thermal resistance to heat transfer of the insulating layer of floors on logs, (m 2 × ° С) / W.

The value is calculated by the expression:

, (2.4)

here is the thermal resistance of closed air layers
(table 2.1); δ d is the thickness of the layer of boards, m; λ d - thermal conductivity of wood material, W / (m · ° С).

Heat loss through the floor located on the ground, W:

, (2.5)

where,,, are the areas of I, II, III, IV zones-stripes, m 2, respectively.

Heat loss through the floor located on the logs, W:

, (2.6)

Example 2.2.

Initial data:

- first floor;

- external walls - two;

- floor construction: concrete floors covered with linoleum;

- design temperature of internal air ° С;

Calculation procedure.

Rice. 2.2. Fragment of the plan and location of floor zones in living room No. 1
(to examples 2.2 and 2.3)

2. Living room no. 1 accommodates only the 1st and part of the 2nd zones.

I-th zone: 2.0´5.0 m and 2.0´3.0 m;

II-nd zone: 1.0´3.0 m.

3. The areas of each zone are equal:

4. Determine the resistance to heat transfer of each zone by the formula (2.2):

(m 2 × ° С) / W,

(m 2 × ° С) / W.

5. Using the formula (2.5), we determine the heat loss through the floor located on the ground:

Example 2.3.

Initial data:

- floor structure: wooden floors on logs;

- external walls - two (Fig. 2.2);

- first floor;

- construction area - Lipetsk;

- design temperature of internal air ° С; ° C.

Calculation procedure.

1. We draw a plan of the first floor on a scale indicating the main dimensions and divide the floor into four zones-stripes 2 m wide parallel to the outer walls.

2. Living room # 1 accommodates only the 1st and part of the 2nd zones.

Determine the size of each strip zone:

To calculate the heat loss through the floor and ceiling, you need the following data:

• the dimensions of the house are 6 x 6 meters.
• The floors are edged boards, grooved with a thickness of 32 mm, sheathed with chipboard 0.01 m thick, insulated with mineral wool insulation 0.05 m thick. Under the house there is an underground for storing vegetables and conservation. In winter, the temperature in the underground is + 8 ° C on average.
• Ceiling - the ceilings are made of wooden panels, the ceilings are insulated from the side of the attic with mineral wool insulation layer thickness of 0.15 meters, with a vapor-waterproofing layer. The attic is not insulated.

Calculation of heat loss through the floor

R boards = B / K = 0.032 m / 0.15 W / mK = 0.21 m²x ° C / W, where B is the thickness of the material, K is the coefficient of thermal conductivity.

R dsp = B / K = 0.01m / 0.15W / mK = 0.07m²x ° C / W

R heat insulation = B / K = 0.05 m / 0.039 W / mK = 1.28 m2x ° C / W

The total value of R floor = 0.21 + 0.07 + 1.28 = 1.56 m2x ° C / W

Considering that in the underground the temperature in winter is constantly kept at about + 8 ° C, the dT required for calculating the heat loss is equal to 22-8 = 14 degrees. Now we have all the data for calculating the heat loss through the floor:

Floor Q = SхdT / R = 36 m2х14 degrees / 1.56 m2х ° С / W = 323.07 Wh (0.32 kWh)

Calculation of heat loss through the ceiling

The ceiling area is the same as the floor S ceiling = 36 m 2

When calculating the thermal resistance of the ceiling, we do not take into account wooden panels, because they do not have a tight connection with each other and do not act as a heat insulator. Therefore, the thermal resistance of the ceiling:

R ceiling = R insulation = thickness of insulation 0.15 m / thermal conductivity of insulation 0.039 W / mK = 3.84 m2x ° C / W

We calculate the heat loss through the ceiling:

Q of the ceiling = SхdT / R = 36 m2х52 degrees / 3.84 m2х ° С / W = 487.5 Wh (0.49 kWh)

Transferring heat through the fences of a home is a complex process. In order to take into account these difficulties as much as possible, the measurement of premises when calculating heat loss is done according to certain rules, which provide for a conditional increase or decrease in area. Below are the main provisions of these rules.

Rules for measuring the areas of enclosing structures: a - section of a building with an attic floor; b - section of a building with a combined coating; в - building plan; 1 - floor above the basement; 2 - floor on logs; 3 - floor on the ground;

The area of ​​windows, doors and other openings is measured by the smallest building opening.

The area of ​​the ceiling (pt) and floor (pl) (except for the floor on the ground) is measured between the axes of the inner walls and the inner surface of the outer wall.

The dimensions of the outer walls are taken horizontally along the outer perimeter between the axes of the inner walls and the outer corner of the wall, and in height - on all floors, except for the lower one: from the level of the finished floor to the floor of the next floor. On the top floor the top of the outer wall coincides with the top of the covering, or attic floor... On the ground floor, depending on the construction of the floor: a) from the inner surface of the floor along the ground; b) from the preparation surface under the floor structure on the logs; c) from the bottom edge of the ceiling over an unheated underground or basement.

When determining heat loss through interior walls their areas are measured along the inner perimeter. Heat losses through the internal fences of the premises can be ignored if the difference in air temperatures in these rooms is 3 ° C or less.

Breakdown of the floor surface (a) and the recessed parts of the outer walls (b) into design zones I-IV

The transfer of heat from a room through the structure of the floor or walls and the thickness of the soil with which they are in contact obeys complex laws. To calculate the resistance to heat transfer of structures located on the ground, a simplified method is used. The surface of the floor and walls (in this case, the floor is considered as a continuation of the wall) along the ground is divided into strips 2 m wide, parallel to the junction of the outer wall and the ground surface.

The counting of zones begins along the wall from the ground level, and if there are no walls along the ground, then zone I is the floor strip closest to outside wall... The next two lanes will be numbered II and III, and the rest of the floor will be zone IV. Moreover, one zone can start on the wall and continue on the floor.

A floor or wall that does not contain insulating layers made of materials with a thermal conductivity coefficient of less than 1.2 W / (m · ° C) are called non-insulated. The heat transfer resistance of such a floor is usually denoted by R np, m 2 ° C / W. For each zone of the non-insulated floor, there are provided normative values heat transfer resistance:

• zone I - RI = 2.1 m 2 ° C / W;
• zone II - RII = 4.3 m 2 ° C / W;
• zone III - RIII = 8.6 m 2 ° С / W;
• zone IV - RIV = 14.2 m 2 ° С / W.

If there are insulating layers in the floor structure located on the ground, it is called insulated, and its resistance to heat transfer R pack, m 2 ° C / W, is determined by the formula:

R pack = R np + R us1 + R us2 ... + R usn

Where R np is the resistance to heat transfer of the considered area of ​​the non-insulated floor, m 2 ° C / W;
R us - resistance to heat transfer of the insulating layer, m 2 · ° C / W;

For a floor on logs, the resistance to heat transfer Rl, m 2 · ° C / W, is calculated by the formula.

Previously, we calculated the heat loss of the floor over the ground for a 6m wide house with a groundwater level of 6m and +3 degrees in depth.
Results and problem statement here -
We also took into account the heat loss to the street air and deep into the earth. Now I will separate the flies from the cutlets, namely, I will carry out the calculation purely into the ground, excluding the heat transfer to the outside air.

I will carry out calculations for option 1 from the previous calculation (without insulation). and the following data combinations
1.GLV 6m, +3 at GWL
2.GLV 6m, +6 on GWL
3. GWL 4m, +3 on GWL
4. GWL 10m, +3 on GWL.
5. GWL 20m, +3 on GWL.
Thus, we will close the issues related to the influence of the depth of the groundwater level and the effect of temperature on the groundwater level.
The calculation, as before, is stationary, does not take into account seasonal fluctuations and does not take into account the outside air at all
The conditions are the same. The soil has Lambda = 1, walls 310mm Lambda = 0.15, floor 250mm Lambda = 1.2.

The results, as before, are two pictures each (isotherms and "IK"), and numerical - the resistance to heat transfer to the ground.

Numerical results:
1.R = 4.01
2.R = 4.01 (Everything is normalized for the difference, otherwise it should not have been)
3.R = 3.12
4.R = 5.68
5.R = 6.14

About the values. If we correlate them with the GWL depth, we get the following
4m. R / L = 0.78
6m. R / L = 0.67
10m. R / L = 0.57
20m. R / L = 0.31
R / L would be equal to one (or rather, the inverse coefficient of thermal conductivity of the soil) for infinitely big house, in our case, the dimensions of the house are comparable to the depth to which heat losses are carried out and what smaller house in comparison with the depth, the less this ratio should be.

The resulting R / L dependence should depend on the ratio of the width of the house to the GWL (B / L), plus, as already mentioned, for B / L-> infinity R / L-> 1 / Lambda.
In total, there are the following points for an infinitely long house:
L / B | R * Lambda / L
0 | 1
0,67 | 0,78
1 | 0,67
1,67 | 0,57
3,33 | 0,31
This dependence is well approximated by the exponential one (see the graph in the commentary).
Moreover, the exponent can be written in a simpler manner without much loss of accuracy, namely
R * Lambda / L = EXP (-L / (3B))
This formula at the same points gives the following results:
0 | 1
0,67 | 0,80
1 | 0,72
1,67 | 0,58
3,33 | 0,33
Those. error within 10%, i.e. very satisfactory.

Hence, for an infinite house of any width and for any GWL in the considered range, we have a formula for calculating the resistance to heat transfer in GWL:
R = (L / Lambda) * EXP (-L / (3B))
here L is the depth of the groundwater level, Lambda is the thermal conductivity of the soil, B is the width of the house.
The formula is applicable in the L / 3B range from 1.5 to about infinity (high GWL).

If we use the formula for deeper groundwater levels, then the formula gives a significant error, for example, for 50m depth and 6m width of the house, we have: R = (50/1) * exp (-50/18) = 3.1, which is obviously too small.

Have a great day everyone!

Conclusions:
1. An increase in the GWW depth does not lead to a corresponding decrease in heat loss in groundwater since everything is involved large quantity soil.
2. At the same time, systems with a GWL of type 20m and more may never go to the hospital received in the calculation during the "life" of the house.
3. R ​​into the ground is not so great, it is at the level of 3-6, so the heat loss deep into the floor along the ground is very significant. This is consistent with the previously obtained result about the absence of a large reduction in heat loss when insulating a tape or blind area.
4. A formula has been derived from the results, use it for health (at your own peril and risk, naturally, I ask you to know in advance that I am not responsible for the reliability of the formula and other results and their applicability in practice).
5. Follows from a little research carried out below in the commentary. Heat loss outside reduces heat loss to the ground. Those. It is incorrect to consider the two heat transfer processes separately. And by increasing heat protection from the street, we increase heat loss to the ground and thus it becomes clear why the effect of warming the house contour previously obtained is not so significant.

The essence of thermal calculations of premises, in varying degrees, located in the ground, is reduced to determining the influence of atmospheric "cold" on their thermal regime, or rather, to what extent a certain soil insulates a given room from atmospheric temperature effects. Because the thermal insulation properties of the soil depend on too many factors, then the so-called 4-zone technique was adopted. It is based on the simple assumption that the thicker the soil layer, the higher its thermal insulation properties (to a greater extent, the effect of the atmosphere is reduced). The shortest distance (vertically or horizontally) to the atmosphere is divided into 4 zones, 3 of which have a width (if it is a floor along the ground) or a depth (if these are walls along the ground) of 2 meters, and the fourth has these characteristics equal to infinity. Each of the 4 zones is assigned its own permanent heat-insulating properties according to the principle - the further the zone (the larger it is serial number), the less the influence of the atmosphere. Omitting the formalized approach, we can make a simple conclusion that the farther a point in the room is from the atmosphere (with a multiplicity of 2 m), the more favorable conditions(from the point of view of the influence of the atmosphere) it will be located.

Thus, the counting of conditional zones begins along the wall from the ground level, provided there are walls along the ground. If there are no walls along the ground, then the first zone will be the floor strip closest to the outer wall. Further, zones 2 and 3 are numbered 2 meters wide. The remaining zone is zone 4.

It is important to consider that the zone can start on the wall and end on the floor. In this case, you should be especially careful when making calculations.

If the floor is not insulated, then the values ​​of the heat transfer resistances of the non-insulated floor by zones are:

zone 1 - R n.p. = 2.1 m2 * C / W

zone 2 - R n.p. = 4.3 m2 * C / W

zone 3 - R n.p. = 8.6 m2 * C / W

zone 4 - R n.p. = 14.2 m2 * C / W

To calculate the resistance to heat transfer for insulated floors, you can use the following formula:

- resistance to heat transfer of each zone of the non-insulated floor, m2 * C / W;

- insulation thickness, m;

- coefficient of thermal conductivity of the insulation, W / (m * C);